The specific heat of the metal object with a mass of 22.7g heated to to temperature of 97.0°C and then transferred to an insulated container containing 84.7 g of water at 20.5 ∘C is 0.815J/g°C
How to calculate specific heat?
The specific heat capacity of a metal can be calculated using the calorimetry equation as follows:
Q = mc∆T
Where;
Q = quantity of heat absorbed
m = mass of substance
c = specific heat capacity
∆T = change in temperature
mc∆T (water) = -mc∆T (metal)
84.7 × 4.18 × 3.8 = - (22.7 × c × -72.7)
1345.375 = 1650.29c
c = 0.815J/g°C
Therefore, the specific heat of the metal object with a mass of 22.7g heated to to temperature of 97.0°C and then transferred to an insulated container containing 84.7 g of water at 20.5 ∘C is 0.815J/g°C.
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Answer:
Explanation:
<em>Percent ionization</em> is the percent of the original acid that has ionized:
- %, ionization = (molar concentration of hydrogen ions at equilibrium / molar concentration of original acid) × 100
<u><em>Part A:</em></u>
<u>1) Data:</u>
- Ka: 6.7 × 10 ⁻⁷
- [HA] = 0.10 M
- %, ionization = ?
<u>2) Equilibrium equation:</u>
<u>3) ICE (initial, change, equilbirium) table </u>
Concentrations
HA H⁺ A⁻
Initial 0.10 0 0
Change - x + x + x
Equilibrium 0.10 - x x x
- Equation: Ka = [H⁺] [A⁻] / [HA] =
6.7 × 10 ⁻⁷ = x² / (0.10 - x)
<u>4) Solve the equation:</u>
Since Ka << 1, you can assume x << 0.10 and 0.10 - x ≈ 0.10
- 6.7 × 10 ⁻⁷ ≈ x² / 0.10 ⇒ x² ≈ 6.7 × 10⁻⁸ ⇒ x ≈ 2.588 × 10⁻⁴
- % ionization ≈ (2.588 × 10⁻⁴ M / 0.1 M) × 100 ≈ 0.2588 % ≈ 0.26% (two significant figures)
<u><em>Part B:</em></u>
<u>1) Data:</u>
- Ka: 6.7 × 10 ⁻⁷
- [HA] = 0.010 M
- %, ionization = ?
<u>2) Equilibrium equation:</u>
<u>3) ICE table:</u>
Concentrations
HA H⁺ A⁻
Initial 0.010 0 0
Change - x + x + x
Equilibrium 0.010 - x x x
- Equation: Ka = [H⁺] [A⁻] / [HA] =
6.7 × 10 ⁻⁷ = x² / (0.010 - x)
<u>4) Solve the equation</u>:
Since Ka << 1, you can assume x << 0.010 and 0.010 - x ≈ 0.010
- 6.7 × 10 ⁻⁷ ≈ x² / 0.010 ⇒ x² ≈ 6.7 × 10⁻⁹ ⇒ x ≈ 8.185 × 10⁻5
- % ionization ≈ (8.185 × 10⁻⁵ M / 0.010 M) × 100 ≈ 0.8185 % ≈ 0.82% (two significant figures)
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The theory of plate tectonics states that the Earth's solid outer crust, the lithosphere, is separated into plates that move over the asthenosphere, the molten upper portion of the mantle. Oceanic and continental plates come together, spread apart, and interact at boundaries all over the planet.
0.00860 km in to scientific notation is 8.60 x10^-3 Km
<em><u>Explanation</u></em>
step for writing scientific notation
- move the decimal place to the right to create a new number from 1 to 10 <em>for example 0.0086 is moved three time to create 8.60</em>
- Determine the exponent that is the number of time you have moved the decimal. <em> </em><em>the exponent for 0.00860 is -3 since the decimal has been moved toward the right</em>
- Put the number into correct form for scientific notation that is <em>8.60 x10^-3</em>